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# Low Pass filter

Discussion in 'Homework Help' started by MilK, Mar 1, 2008.

1. ### MilK Thread Starter Member

Mar 1, 2008
25
0
Hi,

Eqn for low pass RC filter

(Vout/Vin) = (1/jwc) / ((1/jwc)+R)

R = 1khz
C=3.1847 x 10^-8
f= 50khz
Vin= 2v

Find Vout express in Db

My working:

i find the magnitude of both side which gives

vout = 2/ sqrt(1+R^2 w^2 c^2)
sub in all values i get vout = 4.997 x 10^-5

change to db i get : 20log(4.997x10^-5) = -86.02db which is wrong

what when wrong?

Thks.

to admin: Sorry for the "Urgent help needed" topic......pls remove it

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,074
1,776
I don't think the answer is around 5 dB.

I would think Vo = (Vin * Xc) / Z

3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
This is wrong. How did you get this? Show your steps.

This is also wrong. Where did you get it?

4. ### MilK Thread Starter Member

Mar 1, 2008
25
0
I've solve it Thanks

5. ### rwmoekoe Active Member

Mar 1, 2007
172
0
db = 20 log( vo/vi )
= 20 log( zc / (zc+r) )
= 20 log( (1/wc) / (1/wc+r) )
= 20 log( (1 / (2 pi 50k 3.18e-8) / (1 / (2 pi 50k 3.18e-8) + 1k) )
= 20 log( 100 / (100+1k) )
= -20 db

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
This calculation actually comes out to -20.83dB, which is wrong.

You forgot that the capacitor's reactance is 90 degrees out of phase with the resistance. You have to find the vector magnitude of the denominator.
Gain = 20 log( (1/wc) / sqrt((R^2) + (1/wc)^2)
Gain = -20.05dB

The OP wanted Vout when Vin=2V.
Vout=-14.03dB.

7. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,074
1,776
Ron,

I got 198.9 mV as Vo, which is approximately -20.05 dB with respect to 2 Vi.

I didn't understand your Vo being -14.03 dB

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
The OP wrote:
You got Vout=198.9 mV.
20 log (.1989) = -14.03dB.
You are correct in showing gain as -20.05dB, which will of course be true for any input voltage. I realize that dB is a unitless gain measurement (a ratio), but the OP said
I suppose to be correct, we should say
Vout=-14.03dBV, i.e., relative to 1 volt, which identifies it as an absolute measurement rather than a ratio.

9. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,074
1,776
That is true Ron. Had you stated dBV, there wouldn't be an issue. At least you didn't use dBf as the reference ...

10. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Database Format? Divorced Black Female? Decibels (frequency)? I guess I'm not up-to-date on acronyms.

11. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,074
1,776
femtowatt.

I saw it on an HP signal generator back in the early 80s. Not one of the more popular references.

12. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
That's funny! Definitely engineering humor, though. My wife doesn't think much of any of my jokes or "funny" stories. I guess I won't try that one on her.

13. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,074
1,776
If I didn't have to look it up, I would have never remembered it. It was on the old HP-8640 signal generator series.

And to think 1 uV (50 ohms) is 130 dBf. The stuff people remember is amazing sometimes.